Question

If $p$ is a prime, show that the sum of the $(p-1{)}^{th}$ powers of any $p$ numbers in arithmetical progression, wherein the common difference is not divisible by $p$, is less by $1$ than a multiple of $p$.

Answer 1

It is known that if the p terms of a series in A.P. are divided by p the remainders will be $0,1,2,3,....1,p-1$$

Hence disregarding the order of the terms, the series may be represented by $ap,bp+1,cp+2,dp+3,.....,kp+(p-1);a,b,c,d,......,k$ bring the various quotients. With the exception of the first term all the terms of the series are prime to p; hence by Fermat's theorem, their $(p-1)$ the powers are all of the form $n\left(p\right)+1$, whilst that of the first term is of the form $M\left(p\right)$. Thus the sum of the $(p-1)$th

Powers$=M\left(p\right)+p-1=M\left(p\right)-1$